Let \(\Lambda_*\) be the exterior algebra generated by the symbols \(\theset{dx_i}^n\) over a field \(k\). Show that letting \(d = \cdot \vee dx_1\) yields a chain complex \(0 \into \Lambda^0 \into \Lambda^1 \into \cdots \into \Lambda^n \into 0\) with trivial homology. Compute what happens when \(dx_1\) is replaced with an arbitrary non-zero element in \(\Lambda^1\).
Solution
Goal: Let \(W\) be the \(n\)-dimensional \(k\)-vector space with basis \(\{dx_1,\dots,dx_n\}\) and let \(\Lambda^*=\bigwedge^*W\). For \(\omega\in\Lambda^1=W\), define \(d_\omega:\Lambda^p\to\Lambda^{p+1}\) by \(d_\omega(\alpha)=\alpha\wedge\omega\).
Show that for \(\omega = dx_1\), \((\Lambda^*, d_{dx_1})\) is a chain complex with trivial cohomology (except at \(\Lambda^0\) where \(H^0=0\) if treated as reduced, or check all \(H^p\)).
Compute the cohomology of \((\Lambda^*, d_\omega)\) for an arbitrary non-zero element \(\omega \in \Lambda^1\).
1. Verify that \(d_\omega \circ d_\omega = 0\) for any \(\omega \in \Lambda^1\).
1.1. For any \(\alpha \in \Lambda^p\), \(d_\omega(d_\omega(\alpha)) = (\alpha \wedge \omega) \wedge \omega = \alpha \wedge (\omega \wedge \omega)\).
1.2. Since \(\omega \in \Lambda^1\), \(\omega \wedge \omega = -\omega \wedge \omega = 0\) in the exterior algebra (in characteristic 2, \(\omega \wedge \omega = 0\) by definition of exterior algebra alternating product).
1.3. Thus \(d_\omega^2 = 0\), so \((\Lambda^*, d_\omega)\) is a cochain complex (chain complex indexed cohomologically).
1.4. The computation \(d_\omega(d_\omega(\alpha)) = \alpha \wedge (\omega \wedge \omega) = 0\) uses associativity of the wedge product and the fact that \(\omega \wedge \omega = 0\) for \(\omega \in \Lambda^1\); hence \(d_\omega^2 = 0\).
2. Show that \((\Lambda^*, d_{dx_1})\) has trivial homology / cohomology (\(H^p(\Lambda^*, d_{dx_1}) = 0\) for all \(p\)).
2.1. Write \(W=k\cdot dx_1\oplus W'\), where \(W'=\operatorname{span}_k\{dx_2,\dots,dx_n\}\).
2.2. The exterior algebra decomposes as \(\Lambda^*W\cong\Lambda^*(k\cdot dx_1)\otimes_k\Lambda^*W'\).
2.3. \(\Lambda^*(k \cdot dx_1)\) is 2-dimensional, with \(\Lambda^0(k \cdot dx_1) = k \cdot 1\) and \(\Lambda^1(k \cdot dx_1) = k \cdot dx_1\).
2.4. The differential \(d_{dx_1}\) on \(\Lambda^*(k \cdot dx_1)\) maps \(1 \mapsto dx_1\) (isomorphism) and \(dx_1 \mapsto 0\).
2.5. Thus the complex \(0 \to \Lambda^0(k \cdot dx_1) \xrightarrow{d_{dx_1}} \Lambda^1(k \cdot dx_1) \to 0\) is the exact sequence \(0 \to k \xrightarrow{\cdot 1} k \to 0\), which has \(H^0 = 0\) and \(H^1 = 0\).
2.6. By the Künneth formula for complexes of vector spaces over \(k\), \[H^*(\Lambda^*W,d_{dx_1})\cong H^*(\Lambda^*(k\cdot dx_1),d_{dx_1})\otimes_k\Lambda^*W'=0.\]
2.7. Alternatively, every element is uniquely \(\alpha+\beta\wedge dx_1\) with \(\alpha,\beta\in\Lambda^*W'\). Define \(h:\Lambda^p\to\Lambda^{p-1}\) by \(h(\alpha+\beta\wedge dx_1)=\beta\).Then \((d h + h d)(\alpha + \beta \wedge dx_1) = d(\beta) + h(\alpha \wedge dx_1) = \beta \wedge dx_1 + \alpha = \alpha + \beta \wedge dx_1 = \operatorname{id}\).Since \(\operatorname{id}\) is chain homotopic to 0, all homology groups vanish: \(H^p(\Lambda^*, d_{dx_1}) = 0\) for all \(p \in \{0, \dots, n\}\).
2.8. The Künneth formula (<2>6) and the explicit contracting homotopy (<2>7) are two independent arguments; each shows \(H^p(\Lambda^*, d_{dx_1}) = 0\) for all \(p\).
3. Compute the homology when \(dx_1\) is replaced with an arbitrary nonzero \(\omega\in\Lambda^1=W\).
3.1. Write \(\omega=\sum_{i=1}^n c_i dx_i\) with some \(c_j\ne0\). Since \(k\) is a field, extend \(\omega\) to a basis of \(W\) and choose an automorphism \(T:W\to W\) with \(T(dx_1)=\omega\).
3.2. Functoriality of the exterior algebra gives a graded algebra automorphism \(\bigwedge T:\Lambda^*W\to\Lambda^*W\) satisfying \((\bigwedge T)(dx_1)=\omega\).
3.3. For every \(\alpha\in\Lambda^*W\),\[ (\bigwedge T)(d_{dx_1}\alpha)=(\bigwedge T)(\alpha\wedge dx_1)=(\bigwedge T\alpha)\wedge\omega=d_\omega((\bigwedge T)\alpha). \]Thus \(\bigwedge T\) is an isomorphism of cochain complexes\[ (\Lambda^*W,d_{dx_1})\xrightarrow{\cong}(\Lambda^*W,d_\omega). \]
3.4. Therefore \(H^p(\Lambda^*,d_\omega)=0\) for every \(p\).
By 2 the source complex is acyclic, and 3 gives an isomorphism of complexes, so the target complex is acyclic as well.
4. Q.E.D.
4.1. Steps <1>1–<1>3 show that for any non-zero element \(\omega \in \Lambda^1\), the complex \((\Lambda^*, d_\omega)\) has trivial homology in all degrees.